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java解决单缓冲生产者消费者问题示例
摘要:经典的生产者消费者问题模拟。此程序模拟最简单情形——单缓冲。为模拟实际情况,consumeitem和produceitem时加了延时,可以通...

经典的生产者消费者问题模拟。此程序模拟最简单情形——单缓冲。为模拟实际情况,consume item和produce item时加了延时,可以通过修改延时模拟不同的生成消费速率。

[code]

[/co/**

* single buffer consumer-producer problem.

* by xu(xusiwei1236@163.com).

* */

public class ConsumerProducer {

static Object buffer = null;

static Object mutex = new Object();

static Object condConsumer = new Object();

static Object condProducer = new Object();

public static void main(String[] args) {

Thread producer = new Thread() {

public void run() {

//for(int i=0; i<10; i++) {

for(int i=0; ; i++) {

// produce item.

try {

Thread.sleep(1000);

} catch (InterruptedException e) {

e.printStackTrace();

}

String item = new String("item-" + i);

System.out.println("[producer] produced " + item);

// wait for buffer empty.

synchronized (condProducer) {

while(buffer != null) {

try {

condProducer.wait();

} catch (InterruptedException e) {

e.printStackTrace();

}

}

}

// put item to buffer.

synchronized (mutex) {

buffer = item;

System.out.println("[producer] put " + item + " to buffer.");

}

// notify consumers.

synchronized (condConsumer) {

condConsumer.notify();

}

}

}

};

Thread consumer = new Thread() {

public void run() {

//for(int i=0; i<10; i++) {

for( ; ; ) {

// wait for item come.

synchronized (condConsumer) {

while( buffer == null ) {

try {

condConsumer.wait();

} catch (InterruptedException e) {

e.printStackTrace();

}

}

}

// get item from buffer.

String item = null;

synchronized (mutex) {

item = (String)buffer;

buffer = null;

System.out.println(" [consumer] get " + item + " from buffer.");

}

// consume item.

try {

Thread.sleep(500);

} catch (InterruptedException e) {

e.printStackTrace();

}

System.out.println(" [consumer] comsumed " + item);

// notify producers.

synchronized (condProducer) {

condProducer.notify();

}

}

}

};

consumer.start();

producer.start();

}

}de]

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